I need help proving $operatorname{Im}(ie^{-2t}(cos(2t)+isin(2t))=e^{-2t}cos(2t)$
This is an example from my textbook.
$$operatorname{Im}(ie^{-2t}(cos(2t)+isin(2t))=e^{-2t}cos(2t)$$
I don't understand why the imaginary part of this expression equals $e^{-2t}cos(2t)$
Can anyone clarify this?
complex-numbers
edited Nov 29 at 9:42
Bernard
118k638111
asked Nov 29 at 9:10
Boris Grunwald
1487
add a comment |
This is an example from my textbook.
$$operatorname{Im}(ie^{-2t}(cos(2t)+isin(2t))=e^{-2t}cos(2t)$$
I don't understand why the imaginary part of this expression equals $e^{-2t}cos(2t)$
Can anyone clarify this?
complex-numbers
edited Nov 29 at 9:42
Bernard
118k638111
asked Nov 29 at 9:10
Boris Grunwald
1487
add a comment |
This is an example from my textbook.
$$operatorname{Im}(ie^{-2t}(cos(2t)+isin(2t))=e^{-2t}cos(2t)$$
I don't understand why the imaginary part of this expression equals $e^{-2t}cos(2t)$
Can anyone clarify this?
complex-numbers
edited Nov 29 at 9:42
Bernard
118k638111
asked Nov 29 at 9:10
Boris Grunwald
1487
This is an example from my textbook.
$$operatorname{Im}(ie^{-2t}(cos(2t)+isin(2t))=e^{-2t}cos(2t)$$
I don't understand why the imaginary part of this expression equals $e^{-2t}cos(2t)$
Can anyone clarify this?
complex-numbers
complex-numbers
edited Nov 29 at 9:42
Bernard
118k638111
asked Nov 29 at 9:10
Boris Grunwald
1487
edited Nov 29 at 9:42
Bernard
118k638111
asked Nov 29 at 9:10
Boris Grunwald
1487
edited Nov 29 at 9:42
Bernard
118k638111
edited Nov 29 at 9:42
Bernard
118k638111
edited Nov 29 at 9:42
Bernard
118k638111
118k638111
asked Nov 29 at 9:10
Boris Grunwald
1487
asked Nov 29 at 9:10
Boris Grunwald
1487
asked Nov 29 at 9:10
Boris Grunwald
1487
1487
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Recall that for a complex number $z=a+ib$, where $a,binmathbb{R}$, the number $a$ is said to be the real part and $b$ the imaginary part. In your case,
$$ ie^{-2t}(cos(2t)+isin(2t)) = ie^{-2t}cos(2t) + i^2e^{-2t}sin(2t) = underbrace{-e^{-2t}sin(2t)}_{text{real part}}+icdotunderbrace{e^{-2t}cos(2t)}_{text{imaginary part}}. $$
answered Nov 29 at 9:13
MisterRiemann
5,7291624
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 at 9:14
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 at 9:15
Thanks, I get it now :)
– Boris Grunwald
Nov 29 at 9:17
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018379%2fi-need-help-proving-operatornameimie-2t-cos2ti-sin2t-e-2t-cos%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Recall that for a complex number $z=a+ib$, where $a,binmathbb{R}$, the number $a$ is said to be the real part and $b$ the imaginary part. In your case,
$$ ie^{-2t}(cos(2t)+isin(2t)) = ie^{-2t}cos(2t) + i^2e^{-2t}sin(2t) = underbrace{-e^{-2t}sin(2t)}_{text{real part}}+icdotunderbrace{e^{-2t}cos(2t)}_{text{imaginary part}}. $$
answered Nov 29 at 9:13
MisterRiemann
5,7291624
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 at 9:14
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 at 9:15
Thanks, I get it now :)
– Boris Grunwald
Nov 29 at 9:17
add a comment |
Recall that for a complex number $z=a+ib$, where $a,binmathbb{R}$, the number $a$ is said to be the real part and $b$ the imaginary part. In your case,
$$ ie^{-2t}(cos(2t)+isin(2t)) = ie^{-2t}cos(2t) + i^2e^{-2t}sin(2t) = underbrace{-e^{-2t}sin(2t)}_{text{real part}}+icdotunderbrace{e^{-2t}cos(2t)}_{text{imaginary part}}. $$
answered Nov 29 at 9:13
MisterRiemann
5,7291624
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 at 9:14
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 at 9:15
Thanks, I get it now :)
– Boris Grunwald
Nov 29 at 9:17
add a comment |
Recall that for a complex number $z=a+ib$, where $a,binmathbb{R}$, the number $a$ is said to be the real part and $b$ the imaginary part. In your case,
$$ ie^{-2t}(cos(2t)+isin(2t)) = ie^{-2t}cos(2t) + i^2e^{-2t}sin(2t) = underbrace{-e^{-2t}sin(2t)}_{text{real part}}+icdotunderbrace{e^{-2t}cos(2t)}_{text{imaginary part}}. $$
answered Nov 29 at 9:13
MisterRiemann
5,7291624
Recall that for a complex number $z=a+ib$, where $a,binmathbb{R}$, the number $a$ is said to be the real part and $b$ the imaginary part. In your case,
$$ ie^{-2t}(cos(2t)+isin(2t)) = ie^{-2t}cos(2t) + i^2e^{-2t}sin(2t) = underbrace{-e^{-2t}sin(2t)}_{text{real part}}+icdotunderbrace{e^{-2t}cos(2t)}_{text{imaginary part}}. $$
answered Nov 29 at 9:13
MisterRiemann
5,7291624
edited Nov 29 at 9:14
answered Nov 29 at 9:13
MisterRiemann
5,7291624
answered Nov 29 at 9:13
MisterRiemann
5,7291624
answered Nov 29 at 9:13
MisterRiemann
5,7291624
5,7291624
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 at 9:14
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 at 9:15
Thanks, I get it now :)
– Boris Grunwald
Nov 29 at 9:17
add a comment |
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 at 9:14
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 at 9:15
Thanks, I get it now :)
– Boris Grunwald
Nov 29 at 9:17
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 at 9:14
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 at 9:14
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 at 9:15
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 at 9:15
Thanks, I get it now :)
– Boris Grunwald
Nov 29 at 9:17
Thanks, I get it now :)
– Boris Grunwald
Nov 29 at 9:17
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018379%2fi-need-help-proving-operatornameimie-2t-cos2ti-sin2t-e-2t-cos%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
